1

Let $(M, d)$ be a metric space. I define a translation on $M$ to be a function $f$ from $M$ to $M$ such that $d(x,f(x))=d(y,f(y))$ for all $x$ and $y$ in $M$. In a previous question, I asked if every translation was an isometry in the post Are all metric translations isometries. The answer was no, even with continuity, and even with bijectiveness. But what if we have both bijectiveness AND continuity? Is that sufficient to force the translation to be an isometry?

Chill2Macht
  • 20,920
user107952
  • 20,508

1 Answers1

4

Let $M = \{-1, 0, 1, 2\}$ with the standard distance (in $\mathbb R$). Let

$$f(-1) =0\ , f(0) = -1\ , f(1) = 2\ , f(2) =1$$

Then $d(x, f(x)) = 1$ for all $x$ and $4= d(f(0), f(1)) \neq d(0, 1) = 1$